Inequlity

**ns1954** · January 24th 2010, 12:58 PM

Let $a>0$ and $k>1$ . Prove that for comlex $z$

$|arg\ (z)|\leq 2\,\arccos\frac 1k \Rightarrow |a+z|\geq \frac {a+|z|}k$

Hint: $\cos x$ is decreasind in $[o,\pi]$ . Not so difficult, I think?

**halbard** · August 20th 2010, 02:17 AM

How about this?

Let $\theta=\arg z$ . Since $|\arg z|\leq 2\arccos\dfrac1k=2\alpha$ and $\cos x$ is decreasing on $[0,\pi]$ we have

$\cos\theta\geq\cos 2\alpha=2\cos^2\alpha-1=\dfrac2{k^2}-1$ .

Thus

$\mathop{\textrm{Re}} z=|z|\cos\theta\geq\left(\dfrac2{k^2}-1\right)|z|$ .

Now
$<br /> |a+z|^2=a^2+2a\mathop{\textrm{Re}}z+|z|^2\geq a^2+2a|z|\left(\dfrac2{k^2}-1\right)+|z|^2$

Hence
$<br /> |a+z|^2-\dfrac{(a+|z|)^2}{k^2}\geq\left(1-\dfrac1{k^2}\right)(a^2-2a|z|+|z|^2)$

Since $a^2-2a|z|+|z|^2=(a-|z|)^2\geq 0$ and $k>1$ the result follows.

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